Greg Egan

I am a science fiction writer and computer programmer.
Latest novel: MORPHOTROPHIC.
Latest collection: SLEEP AND THE SOUL.
Web site: gregegan.net

Web site: https://www.gregegan.net
Location: Perth, Western Australia

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gregegansf_at_mathstodon.xyz@momostr.pink

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Last Notes

2025-09-15 22:28:03 UTC

There’s a new Hal Hartley movie, “Where to Land”.
https://thefilmstage.com/where-to-land-review-hal-hartley-returns-with-shaggy-and-spellbinding-drama/

2025-09-15 08:52:02 UTC

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TextEdit, XCode, Mail, Pages and Safari all update their search string after it’s changed in another app.
Like you, I don’t see any sharing of the search string in/out of Finder or Music.

2025-09-13 14:03:22 UTC

If you want to feel good about humanity for 2½ hours, “Thirteen Lives” is a tense, moving, inspiring tale of the 2018 Thai cave rescue, when ingenuity, bravery, cooperation, desperation and luck paid off. The bitter aftertaste is how many children we’ve collectively failed since.

2025-09-02 14:00:40 UTC

A triangle has edges of length 2, 3 and 4.
Pick two points P and Q at random in the interior of the triangle, and draw the line L that contains both P and Q.
Which two edges of the triangle is the line L most likely to intersect?

2025-08-26 03:28:14 UTC

One formula to rule them all!
P(d, F) = 4d / ((d+1) F)
In d-dimensional space, d≥2, given a regular polytope with F faces that come in opposite pairs, if you pick two points at random from the interior of the polytope and draw the line that contains them, the probability that the line will intersect both faces of any of the F/2 opposite pairs is P(d, F).
For d=2, this applies to all polygons with an even number of edges.
For d=3, the four Platonic solids apart from the tetrahedron.
For d=4, the hypercube, 16-cell, 24-cell, 120-cell and 600-cell.
For d>4, the hypercubes and the cross-polytopes.

2025-08-25 12:56:21 UTC

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“For even F” should be “For F greater than 4, i.e. excluding the tetrahedron, which doesn’t have pairs of opposite faces.”
I did have to do it case by case! When I get a chance I’ll look more closely at the calculations and see if I can isolate the underlying reason.

2025-08-25 08:51:42 UTC

∑ᵢ₌₀ⁿ⁻²∑ⱼ₌ᵢ₊₁ⁿ⁻¹𝑠𝑖𝑛((𝑗−𝑖)π/𝑛)⁴=3𝑛²/16
Well, obviously.

2025-08-24 12:30:06 UTC

Pick two points uniformly at random inside a regular n-gon.
For even n≥4, the probability that the line containing them will pass through two opposite sides of the n-gon is:
8/(3n)
For odd or even n≥3, the probability that the line will pass through two adjacent sides is:
16 sin(π/n)^4 /(3n)

2025-08-24 08:32:20 UTC

Robert Musil, “A Man Without Qualities”
https://media.mathstodon.xyz/media_attachments/files/115/082/806/326/718/346/original/26bc0ba7909f3ed3.jpg

2025-08-23 22:01:22 UTC

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The solution for d=2 is given here:
https://mathstodon.xyz/@gregeganSF/115076414261569820
The general solution can be found the same way, by noting that the conditions on x₁ and x₂ for fixed values of y₁ and y₂ apply independently to all of the d-1 other coordinates. So the probability we used for d=2, |y₁-y₂|, just becomes |y₁-y₂|^{d-1}.
We then integrate |y₁-y₂|^{d-1} over the square 0 ≤ y₁, y₂ ≤ 1, or double the integral over the triangle where y₁ ≥ y₂. This can be done with a change of variables to:
m = y₁-y₂
p = y₁+y₂
y₁ = (p+m)/2
y₂ = (p-m)/2
The change of variables gives a factor of 1/2, and we integrate:
(1/2) m^{d-1} for p from p=m (i.e. y₂=0) to p=2-m (i.e. y₁=1).
This gives us (1-m) m^{d-1}, which we then integrate for m from 0 to 1, to get:
1/[d(d+1)]
We double to get the integral over the full square, then multiply by d to account for all pairs of opposite sides. So:
P(d) = 2/(d+1)
and:
P(17) = 1/9

2025-08-23 11:27:39 UTC

Here’s some Gemini gaslighting:
• giving a wrong answer to a puzzle
• giving Python code that could test its claim
• asserting it obtained results from the code supporting its claim
• but when I ran the code myself, it showed the claim was false.
https://g.co/gemini/share/2188cebf58cc

2025-08-23 10:32:09 UTC

You wake up in a universe with a different number of spatial dimensions.
Your senses are confused, and you’re too embarrassed to ask someone on the street “Hey, how many dimensions does space have?”
But then you chance upon someone who likes mathematical puzzles, and ask them, “If I picked two points at random inside a hypercubic crate, what would the probability be that the line joining them intersected two opposite sides of the crate?”
They scribble on some paper for a while, then reply “One ninth.”
How many dimensions does space have here?

2025-08-22 07:37:21 UTC

A letter from my local council announcing plans for a “Montorsoli school” made me wonder if it was a typo for “Montessori” … but on closer scrutiny, a religious order really is planning to name this facility after the 16th century Florentine sculptor & friar Giovanni Montorsoli.

2025-08-16 11:23:52 UTC

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I love the fact that 106 years ago Eddington struggled to detect a minuscule amount of gravitational lensing by the sun during an eclipse, but these days every tenth Hubble image of a massive galaxy needs a footnote saying which parts of the picture are the actually galaxy and which are the hugely magnified image of another, far more distant galaxy that just happens to lie behind it.
When I travel back in time to visit Einstein on his death bed and console him about his unified theory not working out, I hope to cheer him up with a collection of Einstein rings, along with a compendium of gravitational wave recordings.

2025-08-09 16:03:47 UTC

Me: Ha, take that, complicated blue curve! I have enclosed you in a circle, and now all my calculations putting bounds on your behaviour will be vastly simpler!
Complicated blue curve: Have you actually proved that the circle encloses me?
Me: I hate you very much.
https://media.mathstodon.xyz/media_attachments/files/114/999/638/506/644/354/original/8ee26ed932f7d7a6.png

2025-05-07 05:38:34 UTC

My 1995 novel DISTRESS — about Theories of Everything, Anthrocosmologists, and artificial islands — was published in French in 1997.
Now le Bélial’ have brought it back into print in a revised edition!
Translated by Bernard Sigaud, cover by Aurélien Police.
https://belial.fr/legacy/a/greg-egan/l-enigme-de-l-univers_belial
English-language editions:
https://www.gregegan.net/BIBLIOGRAPHY/Bibliography.html#p53
https://media.mathstodon.xyz/media_attachments/files/114/464/931/857/325/253/original/ff54716d06497ba0.jpg

2025-05-05 11:27:07 UTC

Here’s a fun puzzle about tidal forces.
Suppose a planet in a tidally-locked circular orbit of radius R around its star experiences tidal stretching T (measured in your favourite units for acceleration per distance).
If a wandering planet fell into this solar system on a head-on collision course for the star, what tidal stretching would it experience when its distance from the star was also R?

2025-05-04 14:14:55 UTC

In base 10, there’s an integer whose reciprocal has a repeating block with all 10 digits exactly once:
1/72,728 = 0.000(0137498625) ...
I found examples for all even bases from 2 to 34 except 8, 16 and 32. And for a few days I thought “If these are the only examples, surely it can’t be hard to prove there are no others.”
But then I stepped back and looked at the simpler things in number theory people have been trying to prove for centuries.
For example: are 3, 5, 17, 257 and 65537 the only primes of the form 2^(2^n)+1?
https://en.wikipedia.org/wiki/Fermat_number
If Euler couldn’t prove this, and nobody since has managed to do it either, it seems very unlikely that I’ll ever prove a conjecture about the bases in which the reciprocal of an integer has a non-redundant pandigital reptend.
But check out base 34!
https://gregegan.net/SCIENCE/Reptends/Reptends.html
https://media.mathstodon.xyz/media_attachments/files/114/449/976/483/147/941/original/b6d0d08b7ea93622.png

2025-05-02 07:45:57 UTC

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I must confess I’m having some trouble figuring out precisely what the puzzle is asking.
I guess it is saying that there are automorphisms of the linked-keyring system that map any pair of rings to any other pair, but this is not true of triples ... but are these automorphisms purely topological bijections on the complement of the rings in R^3, or is there any geometric restriction as well? The bit about not telling left from right suggests that reflections would be valid maps, but I’m a bit nervous about the rings having a specific size and a rigid shape, as if that makes some kind of difference from the case where they were just rubber bands. Or maybe that’s a red herring.

2025-05-02 04:16:11 UTC

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They might need to escalate the prize beyond Pizza Hut gift cards.

2025-05-01 07:00:11 UTC

[Serious documentary voice]: “Observers will be paying close attention to the colour of the smoke emerging from the Sistine Chapel next week, as the conclave follows a time-honoured tradition: blue smoke means the new Pope will be a boy, while pink smoke indicates a girl.”

2025-03-26 22:18:00 UTC

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Yes, I phrased that badly; I knew the other two conditions followed from the first, so I should have said that explicitly, rather than making it sound like they were independent requirements.

2025-03-25 08:10:40 UTC

“… by focusing on tasks that are relatively easy for humans, yet hard, or impossible, for AI, we shine a spotlight on capability gaps that do not spontaneously emerge from "scaling up".”
https://arcprize.org/blog/announcing-arc-agi-2-and-arc-prize-2025

2025-03-20 08:46:56 UTC

This makes me feel very, very old, though maybe in a good way: I lived long enough to see the universe change from “Hmm, maybe heading for a Big Crunch” to “More likely everything receding into the distance behind cosmic horizons” to “WTF, honestly, who knows now?”
https://www.science.org/content/article/mystery-force-behind-universe-s-accelerating-expansion-may-not-be-so-constant-after-all

2025-03-18 13:48:11 UTC

INTERIOR. OVAL OFFICE. January 2033.
Camera tracks across the floor, then ascends to show the surface of the Resolute Desk, where two robotic autopens are fighting, signing various documents then crossing out each other’s signature, while emitting noises like angry steam irons.

2025-03-16 22:50:02 UTC

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The crucial properties that D(a) needs to satisfy in order to get an automorphism are:
D(a c) = a D(c) + D(a) = c D(a) + D(c)
D(1) = 0
while D(a) = 0 for all a is ruled out, because then the automorphism is inner.

2025-03-16 05:20:06 UTC

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I think I’ve found a class-preserving outer automorphism of Aff(ℤ/8)!
You map:
x ↦ a x + b
to:
x ↦ a x + (b + D(a))
where
D(1) = D(3) = 0
D(5) = D(7) = 4
I don’t think this is unique. But I’m still searching through the (4!)^8 choices of permutations of the odd and even parts of ℤ/8, for each value of a.
Conjugation preserves the parity of b, so rather than there being 8! choices of permutations of ℤ/8 for each value of a, there are (4!)^2.

2025-03-15 23:03:42 UTC

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I checked all 4^4 possibilities, but I can’t find any functions where this is true other than the constant functions.
So either my calculations are flawed, or something trickier is happening to b that can't be achieved with multiplication.
Checking all (8!)^4 choices of four independent permutations of ℤ/8 isn’t practical ... and though I *think* the a=1 case must correspond to multiplication by 3, 5 or 7, even 3 * (8!)^3 is too big.

2025-03-15 12:03:28 UTC

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I tried to solve the puzzle, but I must be confused about something.
If you conjugate (a,b) with (c,d) I think you get (a, b c + d - a d), but most importantly the first element, a, is unchanged. So any automorphism of the kind the puzzle is describing must leave the a in x ↦ a x + b unchanged.
Suppose we try to find such an automorphism by brute force, and start by looking at what it does to the subgroup consisting of elements of the form x ↦ x + b (which it must preserve). All it can do is change b by permuting the elements of ℤ/8.
When I checked all 8! permutations of ℤ/8, the only nontrivial ones for which this is an automorphism are those that correspond to multiplication of b by 3, 5 or 7. But these are inner automorphisms, corresponding to conjugation by (3,0), (5,0) or (7,0).
I must be doing something wrong! Can you give me a hint as to where I’ve messed up?

2025-03-15 08:59:45 UTC

Fridge: Now that the weather’s about 10 degrees cooler, I’ll make everything on my shelves about 10 degrees cooler!
Me: What? No!
Fridge: If you don’t like it, *you* need to turn down the cooling with that dial thingy.
Me: Maybe you need to look up the meaning of “thermostat.”

2025-03-07 12:06:21 UTC

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I wonder if the maths adviser for this series, who provided lots of authentically mathy things for them to write on whiteboards, was ever tempted to ask the screenwriter “You do know that having a faster source of primes to use as trial divisors would not help much to crack RSA?”

2025-03-05 13:17:49 UTC

Oh, FFS. If you thought Google Search could be bad these days, pity the people who have habitualised themselves to trying to get urgent information out of Facebook.
https://www.abc.net.au/news/2025-03-05/meta-blocked-facebook-searches-cyclone-alfred/105013014

2025-03-03 12:06:37 UTC

As I continue to hate-watch “Prime Target” out of morbid fascination, I’m reminded that if you want to watch something with the same absurd McGuffin that’s a thousand times better written, and more entertaining on every level, there’s always “Sneakers” (1992).

2025-03-02 07:08:35 UTC

TIL that the function:
f(x) = √[x (a-x)] + √[x (b-x)]
has a maximum where x is half the harmonic mean of a and b:
x_m = a b / (a+b)
where its value is the geometric mean of a and b:
f(x_m) = √[a b]
But this doesn’t generalise if you add more terms, like √[x (c-x)]

2025-02-26 12:27:18 UTC

Alice, Bob, and N people too numerous to name are all given independent, uniformly chosen random numbers between 0 and 1.
Assuming odd N, what is lim N→∞ of the probability that Bob’s number and a majority of the N numbers lie on the same side (i.e. < or >) of Alice’s number.

2025-02-26 08:54:06 UTC

Will I make it through the 3 remaining episodes in time, or will I collide with an immovable alien artefact at a forbidden speed and end up as a bloody decapitated corpse?
(Spoiler: I’ll probably watch all 3 tonight to avoid the risk of cutting it too fine.)
https://media.mathstodon.xyz/media_attachments/files/114/069/341/443/022/029/original/295292d3864f1acf.png

2025-02-20 08:52:32 UTC

A rigid ring or sphere surrounding a single massive body will not be stable, but in the “restricted 3-body problem” where the ring/sphere is infinitesimally light, and one massive body is orbiting another, there can be some stable configurations.
For example, if the lighter of the two bodies is less than 1/9 the mass of the heavier one, a rigid sphere enclosing the lighter body can be stable.
H/T Shubhendu Trivedi
https://academic.oup.com/mnras/article/537/2/1249/7989465

2025-02-16 07:54:10 UTC

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Are you talking about the irritating ads on threadreaderapp dot com, or the current state of Twitter?

2025-02-16 07:22:21 UTC

The work of a CDC Epidemic Intelligence Service Officer
https://threadreaderapp.com/thread/1890587835184324644.html#google_vignette

2025-02-15 08:16:37 UTC

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One way to get the 4-piece dissection is to tile the plane with squares and equilateral triangles, both with area 1.
Find vectors v, w with |v|=the side length of an equilateral triangle with area 1, |w|=2, and det|v,w|=2
v = (2 / 3^{1/4}, 0)
w = (√[4-√3], 3^{1/4})
If we use these vectors to construct a lattice with either the triangles or the squares, suitably offset, they will divide each other into exactly the same four pieces!
https://media.mathstodon.xyz/media_attachments/files/114/006/909/386/088/629/original/e3b8c467b19d6110.png

2025-02-15 08:15:59 UTC

In 1907, Henry Dudeney showed that you could dissect a square into 4 pieces that can be rearranged to form an equilateral triangle.
In 2024, Erik D. Demaine, Tonan Kamata and Ryuhei Uehara finally proved that this cannot be done with *less than* 4 pieces!
https://arxiv.org/abs/2412.03865
(H/T @npub13dp…f0qj )
https://media.mathstodon.xyz/media_attachments/files/114/006/904/936/533/557/original/372fc2028d21d329.mp4

2025-02-11 11:19:04 UTC

Why are Einstein rings so big?
Black holes are ridiculously small, for their mass: the event horizon of a solar-mass black hole has the same circumference as a circle of radius 3 km, and the black disk you would see if you looked towards such a hole, marking the circle where all light ends up being captured, is only 50% wider. From a distance of one astronomical unit, the angular diameter of that black disk would be about 0.01 arc-seconds, which is about one tenth the resolution of the Hubble telescope.
However, if a distant star happened to pass precisely behind such a black hole, the ring of light from the gravitational lensing of the star’s image would be large enough to see with the naked-eye!
This might seem surprising at first, given that the scale of everything about a black hole is set by the Schwarzschild radius. But a simple calculation shows why this still allows an Einstein ring to be far larger than the black disk of the hole itself.
When a ray of light approaches a black hole with an impact parameter (the distance of its asymptotic straight-line path from the centre of the hole) of b, the angle by which it is deflected is given by:
θ ≈ 2 Rs/b
where Rs is the Schwarzschild radius. But we also have:
b ≈ d θ
where d is the distance from the black hole to the point at which all the light rays with this impact parameter are focused.
Combining these equations, we have:
θ^2 ≈ 2 Rs/d
So the angular radius of the Einstein ring is the square root of a multiple of the ratio between the Schwarzschild radius and the distance of the observer, while the angular radius of the black disk is simply a multiple of that ratio.
https://media.mathstodon.xyz/media_attachments/files/113/984/975/081/198/588/original/7cd001a3ebf2a5c5.jpeg

2025-02-10 13:54:18 UTC

What does it mean to say that a composite quantum particle containing red, green and blue quarks is “colourless” and does not feel the strong nuclear force?
Every quark has a wave function with 3 components, which we will call red, green and blue. But like the 3 directions in ordinary space, if you add up 3 equal-sized vectors in each of these directions, the sum certainly isn't zero!
But that’s not how colour charge works. When we have more than one particle, we need to form a larger vector space, with 9 directions for 2 particles, and 27 for 3 particles. Each of these 27 directions corresponds to a choice of colour for each of the particles: RRR, RRG, RRB, etc.
When a composite particle contains quarks of all 3 colours, the colourless state is not just one of the six permutations of RGB; rather, it is a sum of all 6, with a sign that depends on whether the permutation is even or odd:
w = (RGB - RBG + GBR - GRB + BRG - BGR)/√6
Why is this “colourless”?
The strong force depends on what happens to a wave function when you multiply its colour vector with a matrix U that belongs to a group called SU(3), which means it is a 3×3 matrix of complex numbers, with a determinant of 1, and its inverse is the matrix you get by taking its transpose and complex conjugate. If a wave function φ is unchanged by this, if Uφ = φ for every U, then φ will not feel the strong force. But the only 3-dimensional vector for which this is true for every U in SU(3) is φ=0.
However, if you have a composite particle with a 27-dimensional wave function, you need to multiply each of the 3 particle’s wave functions with U.
What happens if we do this for our vector w?

2025-02-08 02:59:32 UTC

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Sure, Kurt. Now shut up and take your citizenship test.

2025-02-03 03:46:49 UTC

If you take a tube and join its two ends so that arrows pointing around the tube agree at the join, you get a torus.
But if you join them so the arrows point in opposite directions, the surface that yields is known as Klein’s bottle.
You’ve probably seen 3D versions of Klein’s bottle where the tube pokes back through itself in order to connect up this way. But in 4D space, the surface need not be self-intersecting.
The image shows a projection of a 4D Klein’s bottle into 3D. Here is how the surface was constructed.
Start with the ellipse whose coordinates are:
(cos v, 0, 2 sin v, 0)
where v ranges from 0 to 2π. This is a cross-section of our tube. Now rotate this by an angle of u in the xy plane, and u/2 in the zw plane. The result is:
(cos u cos v, sin u cos v, 2 cos(u/2) sin v, 2 sin(u/2) sin v)
When u reaches 2π, we end up with:
(cos v, 0, –2 sin v, 0)
This gives exactly the same ellipse as we started with, but it is traced out by v running in the opposite direction. So this surface is Klein’s bottle.
It’s not hard to see that this surface never intersects itself. What’s more, its geometry is flat, like a piece of the Euclidean plane.
To see this, consider the vectors we get by taking the derivative wrt u and v.
D_u = (–sin u cos v, cos u cos v, –sin(u/2) sin v, cos(u/2) sin v)
D_v = (–cos u sin v, –sin u sin v, 2 cos(u/2) cos v, 2 sin(u/2) cos v)
These are orthogonal, and the length of D_u is just 1, meaning our rotation sweeps the ellipse by a uniform amount in a direction orthogonal to the ellipse itself. The only difference between the u, v coordinates and x, y coordinates on a plane is that v doesn’t measure distance around the perimeter of the ellipse.
https://media.mathstodon.xyz/media_attachments/files/113/937/879/865/580/491/original/2524dd460e7510ab.mp4

2025-01-30 22:40:45 UTC

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The paper in Nature Astronomy (open access):
https://www.nature.com/articles/s41550-024-02472-9

2025-01-30 11:25:50 UTC

Samples from asteroid Bennu contain A, G, C, T & U nucleotide bases, and 14 of 20 amino acids used by life — but while we use only left-handed versions of these molecules, Bennu has a roughly equal L/R mixture, puncturing theories that the bias on Earth came from an initial cosmic seeding.
https://www.nature.com/articles/d41586-025-00264-3

2025-01-29 07:11:27 UTC

Breaking: Americans to consider constitutional amendment whereby an executive order can only be issued on a given day if you roll two sixes from a standard pair of dice, and a pardon needs to be drawn from a shuffled stack of 52 cards, one of which reads “Go directly to jail.”

2025-01-28 00:09:21 UTC

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I was surprised too, but I have no experience with these measures.
I imported the image into Mathematica, summed the total brightness across all three colour channels at 10,000 points at the centres of the squares in the grid, and normalised to a sum of 1, to get the probabilities p_i.
Then I used:
H = –Σ_i p_i log_2(p_i)
with the summand being 0 when p_i = 0.
I checked the data by turning it back into a grayscale image, and it looks just like the original to my eye.
So if I’ve made a mistake, it’s hard to see what I’ve done wrong.

2025-01-27 23:44:10 UTC

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12.9576 bits, compared to 13.2877 for a uniform distribution.

2025-01-27 23:00:46 UTC

This analysis of the highly non-uniform distribution of 4-digit PINs has no big surprises, but it’s still sobering to see it spelled out in detail.
https://www.abc.net.au/news/2025-01-28/almost-one-in-ten-people-use-the-same-four-digit-pin/103946842
https://media.mathstodon.xyz/media_attachments/files/113/902/801/408/742/319/original/de6b3de92e20f989.png

2025-01-26 03:42:32 UTC

Today I finally got motivated to hunt down and disable the last of the Enhanced Butt-Dialing-Like Catastrophes That Can Only Happen On a Smartphone functions on my Android: “Smart Replies” in the Messages app.
A slip of the finger caused the phone to respond to a long, nuanced text from a friend sharing both good and bad news with the word “Wow” (message composed and sent immediately, with no option to review.)
I long ago disabled unilateral autocorrect / autocomplete, limiting the keyboard to redlining spelling errors and offering a menu of suggested corrections, but I should have realised that this other booby trap was just as dangerous.

2025-01-24 10:52:35 UTC

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It’s worth stressing that when Alice makes her measurement on particles X and A, all four outcomes always have equal probability of 1/4. Despite the fact that the precise state of particle X depends on α and β, those parameters have absolutely no effect on the probability of her finding X and A jointly in each of the \(e_nn\).
So, as far as Alice is concerned, she has just generated a random 2-bit number. But to Bob, it contains precisely what he needs to transform his particle B into a state that contains all the information that Alice previously could have extracted from particle X.

2025-01-24 09:45:31 UTC

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How does this work? You can rewrite \(\psi_{XAB}\) as:
\[\psi_{XAB} = (e_{00} (-\alpha u_B - \beta d_B) +
e_{01} (-\alpha u_B + \beta d_B) +
e_{10} ( \alpha d_B + \beta u_B) +
e_{11} ( \alpha d_B - \beta u_B)) / 2\]
The four states for particle B on the RHS can all be transformed by an
appropriate operation into:
\[\alpha u_B + \beta d_B\]
up to an overall phase, either by doing nothing or by measuring a spin,
i.e. \(U_{nn}\) applied to the term following \(e_{nn}\) will always yield \(\alpha u_B +
\beta d_B\) if you choose:
\[U_{00} = - I\]
\[U_{01} = - \sigma_z\]
\[U_{10} = \sigma_x\]
\[U_{11} = i \sigma_y\]
So after Bob receives a classical two-bit message from Alice telling him which of the \(e_{nn}\) she measured particles X and A to be in, he can then perform the corresponding unitary operation \(U_{nn}\) on particle B ... and the result will be that particle B will end up in the same state as particle X was in initially!
But Bob can't do anything until he receives the message from Alice telling him the result of her measurement. Nothing is "teleported" until he performs the correct, matching operation himself, and he can't know what that is until he has heard from Alice.

2024-11-27 06:13:44 UTC

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The authors of the paper speculate about the possible applications of that.

2024-11-27 05:07:42 UTC

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As they say in the paper, this is still an open question, so maybe you can prove it!

2024-11-27 02:46:47 UTC

In 1926, Menger proved that you can embed any compact 1-dimensional space in his famous fractal sponge, including a loop, but he didn’t say anything about the knottedness of the embedded loop.
Now three teenagers and their mentor have proved that you can actually embed any *knot* in the Menger sponge: you can deform any knot without the string having to pass through itself so that it ends up as a subset of the Menger sponge.
https://www.quantamagazine.org/teen-mathematicians-tie-knots-through-a-mind-blowing-fractal-20241126/
https://arxiv.org/abs/2409.03639
https://media.mathstodon.xyz/media_attachments/files/113/552/596/328/062/847/original/5c2f06a9b5294af3.png

2024-11-25 07:16:30 UTC

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Time reversal symmetry doesn’t say that you can time reverse system A while not time reversing system B, if systems A and B interact. The time reverse of a particle falling into a black hole is not a particle emerging from a black hole, it’s a particle emerging from the time reverse of a black hole.
The time reverse of a black hole would not violate GR per se, but there are other reasons (thermodynamic, cosmological, causal complications) that suggest such things probably don’t exist in the real universe.

2024-11-24 16:05:50 UTC

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OK, I see what’s troubling you.
The horizon isn’t really a “place”: it’s traced out by null rays in spacetime, rather than timelike rays. For a black hole, the horizon has a constant area, which makes it seem like a thing that’s standing still in some sense, but it’s generated, geometrically, by light rays, so it’s moving at the speed of light.
For two people who cross the horizon at different times, the latter one will see the former one when they are both crossing the horizon, but that doesn’t imply that they bump into each other.
Bob’s notion of his distance from *the horizon* (in the sense of the distance to a spacetime event that lies on the horizon “right now”, in his reference frame and with his notion of simultaneity) goes from being fixed to being monotically decreasing when he jumps out of the ship.
But his notion of his distance *from Alice* when she emitted the light with which he is currently seeing her starts *increasing*.
That’s not a contradiction, because these are two different things.

2024-11-24 14:30:35 UTC

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OK, Alice jumps first, followed by Bob.
Bob does not see light from Alice at the moment she crossed the horizon until he, too, is crossing the horizon. He certainly doesn’t see her pinned to the horizon and fading away from redshift, as he would have if he didn’t fall himself and just stayed at a fixed distance from the horizon.
But I don’t know why you think the horizon “recedes”. Alice recedes from Bob, because she fell first. He reaches the horizon himself in a finite time by his own personal clock, and at that point he sees Alice at the moment she crossed the horizon.

2024-11-24 13:56:55 UTC

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If you mean the world lines of infalling dust particles on that Penrose diagram, I’m not sure; I don’t know exactly what coordinate system is used there.
BTW, this other page I wrote on things falling into black holes might also be of interest:
https://www.gregegan.net/SCIENCE/FiniteFall/FiniteFall.html

2024-11-24 04:23:50 UTC

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I did a long set of calculations exploring various scenarios for lowering things through a horizon. These calculations use a Rindler Horizon, but that works as a good approximation to the horizon of a black hole large enough that tidal effects are irrelevant.
It’s not exhaustive, but it’s nice to have some explicit calculations that are near enough to a few of the “Yes, but what if I did X?” questions that people sometimes propose as ways to “cheat” an event horizon.
https://www.gregegan.net/SCIENCE/Rindler/RindlerHorizon.html

2024-11-21 19:23:57 UTC

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And the coolest tidal fact of all (which Rob once prompted me to realise) is that the near-perfect angular size match in solar eclipses means that essentially all the difference in lunar vs solar tidal strength (in the approximation that the separation distances are much larger than the radii of the bodies) is due to the ratio of densities between the sun and the moon.

2024-11-20 15:52:28 UTC

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Me: Please, mathematics, grant me a deep geometric understanding of these equations!
Mathematics: No, but if you tinker with your Mathematica code, you can have smaller factors in the exponentially growing time & space complexity of computing them.
Me: OK, better than nothing.
https://media.mathstodon.xyz/media_attachments/files/113/516/074/138/214/630/original/703202d7833e4a94.png

2024-11-18 12:57:44 UTC

Me after the first three equations: I bet there’s a simple way to generalise these conditions!
After the fourth: Maybe there’s a subtle pattern here … ?
After the fifth: [sound of me banging my head on my desk].
https://media.mathstodon.xyz/media_attachments/files/113/504/064/491/433/332/original/e5d0d47bebd027ad.png

2024-09-12 11:52:29 UTC

Does anyone really watch foreign-language programs dubbed into English, when subtitles are available? I hadn't been hit with the old lips vs dialogue mismatch for about 50 years, but suddenly I’m seeing Apple TV promos as hilarious as those dubbed spaghetti Westerns from the 70s.